Integrand size = 60, antiderivative size = 24 \[ \int -\frac {12 e^{e^{\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}+\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}}{4+e^x} \, dx=3 e^{e^{\frac {\log ^2(10)}{5}} \left (1+4 e^{-x}\right )} \] Output:
3*exp(exp(1/5*ln(10)^2+ln(4/exp(x)+1)))
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int -\frac {12 e^{e^{\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}+\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}}{4+e^x} \, dx=3 e^{e^{-x+\frac {\log ^2(10)}{5}} \left (4+e^x\right )} \] Input:
Integrate[(-12*E^(E^((Log[10]^2 + 5*Log[(4 + E^x)/E^x])/5) + (Log[10]^2 + 5*Log[(4 + E^x)/E^x])/5))/(4 + E^x),x]
Output:
3*E^(E^(-x + Log[10]^2/5)*(4 + E^x))
Time = 0.42 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {27, 2720, 2638}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int -\frac {12 \exp \left (\frac {1}{5} \left (5 \log \left (e^{-x} \left (e^x+4\right )\right )+\log ^2(10)\right )+e^{\frac {1}{5} \left (5 \log \left (e^{-x} \left (e^x+4\right )\right )+\log ^2(10)\right )}\right )}{e^x+4} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -12 \int \frac {\exp \left (e^{\frac {\log ^2(10)}{5}-x} \left (4+e^x\right )+\frac {1}{5} \left (5 \log \left (e^{-x} \left (4+e^x\right )\right )+\log ^2(10)\right )\right )}{4+e^x}dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle -12 \int \exp \left (-2 x+4 e^{\frac {\log ^2(10)}{5}-x}+\frac {1}{5} \left (5 e^{\frac {\log ^2(10)}{5}}+\log ^2(10)\right )\right )de^x\) |
\(\Big \downarrow \) 2638 |
\(\displaystyle 3 e^{4 e^{\frac {\log ^2(10)}{5}-x}+e^{\frac {\log ^2(10)}{5}}}\) |
Input:
Int[(-12*E^(E^((Log[10]^2 + 5*Log[(4 + E^x)/E^x])/5) + (Log[10]^2 + 5*Log[ (4 + E^x)/E^x])/5))/(4 + E^x),x]
Output:
3*E^(E^(Log[10]^2/5) + 4*E^(-x + Log[10]^2/5))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> Simp[(e + f*x)^n*(F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n *Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] && EqQ [d*e - c*f, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Time = 0.47 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
norman | \(3 \,{\mathrm e}^{{\mathrm e}^{\frac {\ln \left (10\right )^{2}}{5}} \left ({\mathrm e}^{x}+4\right ) {\mathrm e}^{-x}}\) | \(20\) |
derivativedivides | \(3 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right )+\frac {\ln \left (10\right )^{2}}{5}}}\) | \(22\) |
default | \(3 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right )+\frac {\ln \left (10\right )^{2}}{5}}}\) | \(22\) |
parallelrisch | \(\frac {3 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right )+\frac {\ln \left (10\right )^{2}}{5}}} {\mathrm e}^{2 x}+24 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right )+\frac {\ln \left (10\right )^{2}}{5}}} {\mathrm e}^{x}+48 \,{\mathrm e}^{{\mathrm e}^{\ln \left ({\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right )+\frac {\ln \left (10\right )^{2}}{5}}}}{\left ({\mathrm e}^{x}+4\right )^{2}}\) | \(79\) |
risch | \(3 \,{\mathrm e}^{2^{\frac {2 \ln \left (5\right )}{5}} \left ({\mathrm e}^{x}+4\right ) {\mathrm e}^{-x +\frac {i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+4\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+4\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )}{2}-\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )}{2}+\frac {\ln \left (2\right )^{2}}{5}+\frac {\ln \left (5\right )^{2}}{5}}}\) | \(134\) |
Input:
int(-12*exp(ln((exp(x)+4)/exp(x))+1/5*ln(10)^2)*exp(exp(ln((exp(x)+4)/exp( x))+1/5*ln(10)^2))/(exp(x)+4),x,method=_RETURNVERBOSE)
Output:
3*exp(exp(1/5*ln(10)^2)*(exp(x)+4)/exp(x))
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (20) = 40\).
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.42 \[ \int -\frac {12 e^{e^{\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}+\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}}{4+e^x} \, dx=\frac {3 \, e^{\left (\frac {1}{5} \, {\left (e^{x} \log \left (10\right )^{2} + 5 \, {\left (e^{x} + 4\right )} e^{\left (\frac {1}{5} \, \log \left (10\right )^{2}\right )} + 5 \, e^{x} \log \left ({\left (e^{x} + 4\right )} e^{\left (-x\right )}\right )\right )} e^{\left (-x\right )} - \frac {1}{5} \, \log \left (10\right )^{2} + x\right )}}{e^{x} + 4} \] Input:
integrate(-12*exp(log((exp(x)+4)/exp(x))+1/5*log(10)^2)*exp(exp(log((exp(x )+4)/exp(x))+1/5*log(10)^2))/(exp(x)+4),x, algorithm="fricas")
Output:
3*e^(1/5*(e^x*log(10)^2 + 5*(e^x + 4)*e^(1/5*log(10)^2) + 5*e^x*log((e^x + 4)*e^(-x)))*e^(-x) - 1/5*log(10)^2 + x)/(e^x + 4)
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int -\frac {12 e^{e^{\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}+\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}}{4+e^x} \, dx=3 e^{\left (e^{x} + 4\right ) e^{- x} e^{\frac {\log {\left (10 \right )}^{2}}{5}}} \] Input:
integrate(-12*exp(ln((exp(x)+4)/exp(x))+1/5*ln(10)**2)*exp(exp(ln((exp(x)+ 4)/exp(x))+1/5*ln(10)**2))/(exp(x)+4),x)
Output:
3*exp((exp(x) + 4)*exp(-x)*exp(log(10)**2/5))
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (20) = 40\).
Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int -\frac {12 e^{e^{\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}+\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}}{4+e^x} \, dx=3 \, e^{\left (2^{\frac {2}{5} \, \log \left (5\right ) + 2} e^{\left (\frac {1}{5} \, \log \left (5\right )^{2} + \frac {1}{5} \, \log \left (2\right )^{2} - x\right )} + 2^{\frac {2}{5} \, \log \left (5\right )} e^{\left (\frac {1}{5} \, \log \left (5\right )^{2} + \frac {1}{5} \, \log \left (2\right )^{2}\right )}\right )} \] Input:
integrate(-12*exp(log((exp(x)+4)/exp(x))+1/5*log(10)^2)*exp(exp(log((exp(x )+4)/exp(x))+1/5*log(10)^2))/(exp(x)+4),x, algorithm="maxima")
Output:
3*e^(2^(2/5*log(5) + 2)*e^(1/5*log(5)^2 + 1/5*log(2)^2 - x) + 2^(2/5*log(5 ))*e^(1/5*log(5)^2 + 1/5*log(2)^2))
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int -\frac {12 e^{e^{\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}+\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}}{4+e^x} \, dx=3 \, e^{\left (e^{\left (\frac {1}{5} \, \log \left (10\right )^{2}\right )} + 4 \, e^{\left (\frac {1}{5} \, \log \left (10\right )^{2} - x\right )}\right )} \] Input:
integrate(-12*exp(log((exp(x)+4)/exp(x))+1/5*log(10)^2)*exp(exp(log((exp(x )+4)/exp(x))+1/5*log(10)^2))/(exp(x)+4),x, algorithm="giac")
Output:
3*e^(e^(1/5*log(10)^2) + 4*e^(1/5*log(10)^2 - x))
Time = 7.44 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int -\frac {12 e^{e^{\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}+\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}}{4+e^x} \, dx=3\,{\mathrm {e}}^{4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{\frac {{\ln \left (10\right )}^2}{5}}}\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\ln \left (10\right )}^2}{5}}} \] Input:
int(-(12*exp(log(exp(-x)*(exp(x) + 4)) + log(10)^2/5)*exp(exp(log(exp(-x)* (exp(x) + 4)) + log(10)^2/5)))/(exp(x) + 4),x)
Output:
3*exp(4*exp(-x)*exp(log(10)^2/5))*exp(exp(log(10)^2/5))
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int -\frac {12 e^{e^{\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}+\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}}{4+e^x} \, dx=3 e^{\frac {e^{\frac {\mathrm {log}\left (10\right )^{2}}{5}+x}+4 e^{\frac {\mathrm {log}\left (10\right )^{2}}{5}}}{e^{x}}} \] Input:
int(-12*exp(log((exp(x)+4)/exp(x))+1/5*log(10)^2)*exp(exp(log((exp(x)+4)/e xp(x))+1/5*log(10)^2))/(exp(x)+4),x)
Output:
3*e**((e**((log(10)**2 + 5*x)/5) + 4*e**(log(10)**2/5))/e**x)